Chapter A.6, Problem 140AYU

To determine

To show: The real solutions of the equation $a{x}^2+ bx + c = 0$ are the reciprocals of the real solutions of the equation $c{x}^2+ bx + a = 0$. Assume that $b^2-4ac\ge 0$.

Answer to Problem 140AYU

The real solutions of the equation $a{x}^2+ bx + c = 0$ are the reciprocals of the real solutions of the equation $c{x}^2+ bx + a = 0$.

Explanation of Solution

The equation is $a{x}^2+ bx + c = 0$.

The roots are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

The equation is $c{x}^2+ bx + a = 0$.

The roots are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2c}$.

Since $b^2-4ac\ge 0$.

$\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\times \left(\frac{-b-\sqrt{b^2-4ac}}{2c}\right)=\frac{{\left(-b\right)}^2-{\left(\sqrt{b^2-4ac}\right)}^2}{4ac}=\frac{b^2-b^2+4ac}{4ac}=\frac{4ac}{4ac}=1$

$\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)\times \left(\frac{-b+\sqrt{b^2-4ac}}{2c}\right)=\frac{{\left(-b\right)}^2-{\left(\sqrt{b^2-4ac}\right)}^2}{4ac}=\frac{b^2-b^2+4ac}{4ac}=\frac{4ac}{4ac}=1$

Therefore the roots are reciprocal, since the product of roots are one.